Mechanical Systems (part 1)

Revised Standard Grade Technological Studies

Mechanical Systems

Standard Grade Technological Studies: Mechanical Systems ii

Contents

Preface iii

Structure iii

Resources iv

Assessment iv

Introduction 1

Motion 2

Rotary 2

Linear 2

Reciprocating 2

Oscillating 2

Forces 5

Static forces 5

Dynamic forces 5

Bending forces 6

Shear forces 6

Torsion forces 6

Compression forces 7

Tension forces 7

Vectors 9

Equilibrium 10

Levers 15

Force multiplier ratio 16

Movement multiplier ratio 16

Efficiency 17

Classes of lever 20

Moments 25

The bell-crank lever 32

Linkages 34

Free-body diagrams 37

Beams 38

Beam reactions 38

Types of beam support 38

Gears 44

Simple gear train 44

Movement multiplier ratio in gears 45

Idler gears 46

Ratchet and pawl 46

Compound gears 50

Worm and wheel 51

Bevel gears 52

Torque and Drive Systems 57

Power transmitted by a belt drive 58

Belt and chain drives 59

Multiplier ratio for belt drives 61

Toothed belts 65

Chain drives 65

Multiplier ratio for chain drives 67

Chain tension 67

Converting motion 71

Cams 71

Cam motion 71

Crank slider 73

Rack and pinion 77

Couplings 81

Bearings 82

Clutches 84

Mechanical Systems – Homework 89

Unit Assessment 113

Standard Grade Technological Studies: Mechanical Systems

iii

Preface

Mechanisms are widely used in industry and are included in a large number of devices

and equipment commonly used in society. It is important, therefore, that students have

a good understanding of how mechanical systems assist and aid everyday tasks.

Throughout the coursework, students should gain knowledge of how mechanical

systems operate and how they fit into common everyday devices. It is not only

necessary to understand how the mechanical system carries out its task, but also

important to know hot the systems can be altered to improve performance and

efficiency. It is therefore recommended that this unit of work be completed towards

the end of the course to allow students to grasp the requirements and rigour of the

mechanisms calculations.

Owing to the nature of course modifications, it is not necessary for students to

construct mechanical or compound systems. If, however, the class teacher sees a need

for construction to assist the learning process, then this can be undertaken. The

construction of ‘rigs’ or working models to display and convey important mechanical

observations is also acceptable.

Structure

The material provides the student with natural progression throughout the course.

There is, however, flexibility within the material so that the teacher can determine

which sections can be taught to suit the classroom needs.

The following areas are covered.

Introduction

Motion

Forces

Vectors

Levers

Moments

Linkages

Free-body diagrams

Beams

Gears

Torque and drive systems

Converting motion

Rack and pinion

Couplings, bearings, clutches,

friction

The content of the unit fulfils all the requirements of the course and the structured

homework exercises can be distributed at the end of a course topic. Some teachers

may, however, add additional well-tried and tested worksheets to benefit student

learning and understanding of topics.

The use of the CD-ROM gives students an interactive way of understanding

mechanical devices in real-life situations. This classroom methodology is a very

important part of students’ development and their knowledge of how mechanical

systems work can assist in everyday situations. The tasks associated with the CD-

ROM allow pupils to search and interact at suitable times within the coursework.

There is a variety of suitable CD-ROMs on mechanisms and these could also be used

at the discretion of the teacher.

The interactive computer simulation associated with the tasks:

Standard Grade Technological Studies: Mechanical Systems iv

6 Levers

4 Gears and Belts

12 Gears and Belts

2 Cams and Cranks

2 Friction

refers to The New Way Things Work CD-ROM.

The ‘Simulation of Gear Systems: Task 1 Simple gear train’ refers to Crocodile

Clips Mechanical components.

Resources

The majority of resources required to assist in the understanding of this unit should be

available if Technological Studies has already been undertaken. The resources

comprise:

examples of classes of levers

a balance lever

a model of a gear train

a model of a compound gear system

a computer simulation of gear systems

an interactive computer simulation of mechanisms

a model of a bevel-gear system

a model of belt/chain-drive systems

a model of a rack-and-pinion system

examples of couplings, bearings and clutches

individual components.

Assessment

Internal assessment

Internal assessment can be measured against the tasks that the students complete

during the coursework allied with their homework at the end of topics.

A twelve-question end-of-unit assessment prepares students to undertake assessable

elements in knowledge and understanding and reasoning and numerical analysis. This

assessable work can be used to supplement students’ grades in the event of appeals.

Marks are awarded for each question and are associated with elements.

External assessment

This unit of work and the exercises within prepare the students for any mechanisms

questions that appear in the 90-minute exam at the end of the course. It will enable all

students to gain the knowledge and understanding required and give them suitable

practice in reasoning and numerical analysis.


1

Introduction

Mechanisms are still a large part of modern society. Most of the mechanisms that we

use every day are so familiar that we never think twice about them, for example door

handles, light switches, scissors, etc.

Mechanisms play a vital role in industry. While many industrial processes now have

electronic control systems, it is still mechanisms that provide the muscle to do the

work. They provided the forces to press steel sheets into the shape of car body panels,

to lift large components from place to place and to force power hacksaws to cut

through thick metal bars – the list of jobs is endless. It is only by using mechanisms

that industry can make products you use every day.

Some machines are easy to understand, but many are hidden away from sight behind

glossy panels and covers. In the past, machines were much easier to see, as with the

old steam engine, for example, but as people became more concerned about safety, it

was necessary to fit guards over moving parts. Today, guards are often replaced by

styled covers that make it much harder to see what is happening, but whether you can

see them or not, mechanisms are still playing a vital part in everyday life.

All mechanisms:

involve some kind of motion

involve some kind of force

make a job easier to do

need some kind of input to make them work

produce some kind of output.

Standard Grade Technological Studies: Mechanical Systems 2

Motion

There are four basic kinds of motion.

Rotary

Turning in a circle. This is the most common

type of movement, for example wheels, clock

hands, compact discs, CD-ROMs.

Linear

Movement in a straight line, for example

movement of a paper trimmer cutting a straight

edge on paper or a lift moving between floors.

Reciprocating

Backwards and forwards movement in a straight

line, for example the needle in a sewing machine or

the piston in a car engine.

Oscillating

Swinging backwards and forwards in an arc, for

example the pendulum of a clock, a playground

swing or a rocking horse.


3

Motion: task 1

What types of motion do the following sports or leisure activities show when they are

being used or carried out? Complete a systems diagram for each.

Swing

100 metres’ sprint

Golfing

Bungee jump

See-saw

Fire button on a computer game


Motion: task 2

The machines and tools that are used in your practical rooms in school use all types of

motion. The four types of motion are listed; now list as many machines/tools as

possible for each type of motion.

Rotary

Linear

Reciprocating

Oscillating


5

Forces

Forces affect structures in a variety of different ways depending on how they are

applied to the structure. Forces can move a structure slightly or cause damage by

changing the shape of the structure.

Sometimes when forces are applied to a structure, it may be almost impossible to see

changes happening. For example, a bridge will sag slightly when a vehicle drives over

it, but this is not visible to the human eye. Nevertheless, the vehicle causes

downwards movement of the bridge structure. Loads such as vehicles on a bridge can

be deemed examples of forces acting on the bridge. Forces can stop an object from

moving or they can make it change direction. When a football is kicked, the forces

applied from the player cause the dimensions of the ball to change on impact. It

happens so quickly that it is not visible.

Forces are measured in newtons and the symbol is the letter ‘N’.

There are a number of different types of forces that can be applied to and which affect

bodies and structures.

Static forces

When static loads or forces are applied to structures, the structures do not normally

move. Normally the total downwards force comprises the weight of the structure plus

the load it is carrying. The runner below is in his starting position; his weight is a

static or stationary downwards force.

Dynamic forces

When dynamic loads or forces are applied to a structure, the structure does move and

the forces applied can be varied. Dynamic forces are visually more noticeable and are

produced by a variety of means and effects: machines, wind directions, people, etc.

The picture below shows the sprinter after the starting gun has been fired; he is

creating a dynamic impact to gain momentum.


Bending forces

Structures that carry loads across their length are subject to bending forces. The

weightlifter lifting a weighted bar feels the effect of the downward forces of the

weights and these cause the bar to bend.

A car driving across a bridge will cause bending forces on the structure but often they

are not visible.

Shear forces

Shear forces can be described as tearing or cutting forces affecting structures. Simple

examples are a pair of scissors used to cut a ribbon at an opening ceremony and a

mower cutting the grass.

Torsion forces

Torsion or torque forces have the effect of trying to turn or twist a structure or a piece

of material. A screwdriver being twisted to apply a force to a screw and a spanner

turning a bolt to lock it into place are examples of torque being applied.


7

COLUMN

WEIGHT FORCE W(EXTERNAL FORCEON COLUMN)

W

R

GROUND REACTION R(EXTERNAL FORCE ON COLUMN)

Compression forces

The figure below shows a column with a weight pressing down on it, but the column

does not disappear into the ground because the ground exerts an upwards reaction

force on the column’s base. The downward pressure of the weight and the upward

reaction are external forces trying to squash or shorten the column. Forces that act like

this are called compressive forces and the column is said to be in compression.

For example, when you sit on a stool in the classroom, your weight acts as a

downward force on the chair. However, there must be an

upward force on the legs of the chair; therefore the legs are

said to be in compression.

The same can be said about the weightlifter’s arms and legs.

Figure 1

Tension force

We have noted that compression occurs when things are being pushed together. The

opposite of compression is ‘tension’ – when a structure is being pulled apart. In a tug

of war, the two sides are pulling the rope in opposite directions. The forces applied by

the teams are called tensile forces and cause the rope to be in tension. It could also be

said that the arms of team members are in tension.

Figure 2

The wire rope holding the net in volleyball is also in

tension.

Figure 3


Force: task 1 Against each of the six forces mentioned make a list of ‘real life’ situations where

these types of forces may be found. Ask the teacher if you are unsure which category

the situations fit into.

(a) Static

(b) Dynamic

(c) Bending

(d) Shear

(e) Torsion

(f) Compression

(g) Tension


9

Vectors

Force is a vector quantity and has both magnitude and direction. This means it is often

convenient to represent a force by a line, that is, a vector quantity, which is sometimes

easier to understand visually. The direction of the force may be indicated by an arrow-

headed line, with the length of the line drawn to scale to represent the size of the

force. This line is called a vector.

Example The cyclist pedalling with a force of 800 N is being assisted by a tail-wind of 400 N,

but the friction from the road surface measures 200 N.

Figure 1

The overall effect will be 800 N + 400 N – 200 N = 1000 N (or 1 kN).

A suitable scale would be selected – possibly 10 mm to represent 20 N – and using

this scale each force is drawn in turn, one following on from the other.

Figure 2: a vector diagram

When the three forces are added together, they can be replaced by a single force that

has the same effect, called the ‘resultant’.

Figure 3

800 N

400 N

200 N

800 N + 400 N – 200 N = 1000 N or 1 kN

RESULTANT = 1 kN


Vectors are also used to find the resultant of two forces that are inclined at an angle to

each other.

Figure 4

In the example above the resultant of the two forces can be found by drawing two

vectors. First choose a suitable scale and draw the two vectors CA and CB.

Scale: 10 mm = 10 N

25 mm = 25 N = CA

35 mm = 35 N = CB

The bigger the scale the more accurate the vectors.

From A draw a line parallel to CB, and from B draw a line parallel to CA. Call the

point where the two lines intersect point D. Now draw a line from C to D. A line

drawn from C to D is the resultant of the two forces CA and CB.

Figure 5

The resultant has a magnitude of 46 N by measurement.

Equilibrium

Certain conditions must apply within structures in order to create stability. The

resultant is made up of the combined forces that are trying to move an object or

structure in a set direction. If such a force were applied without an opposing force

then major problems could occur. Structures have to remain in a stable or balanced

state called ‘equilibrium’, which simply means ‘balanced’. There are three types of

balancing that must exist if structures, bodies, objects, etc. are to remain in

equilibrium: horizontal, vertical and rotational forces must all balance.

60 º

35 N

25 N


11

The general conditions of equilibrium are as follows.

upward forces = downward forces

leftward forces = rightward forces

clockwise moments = anticlockwise moments

Example 1 Consider again the same two forces in figure 5. Are they in equilibrium? It is easily

seen that a force must be added acting downwards to the left, but we cannot tell from

this exactly how large this force must be or its exact direction (figure 6).

Figure 6

The resultant has been drawn and it can be seen that to balance it, the equilibrant CE

is required. If the forces F1 and F2 are drawn as in figure 7 then it is much easier to

obtain the equilibrant by completing the triangle, as shown in figure 7.

Figure 7


Example 2

A crane is fixed against a wall, as shown in figure 8. Find the forces in the

compression and tension members.

Figure 8

To find the forces created in the tension and compression members by the 1000 N

load, draw the triangle from the area circled. Select a suitable scale and then draw the

known force first, the 1000 N load (figure 8). A line is drawn through one end of the

load line parallel to one of the unknown forces. Another line is drawn through the

other end, parallel to the second unknown force. By measuring each line, the size of

each force can be found. (Note: the arrowheads must follow round the triangle.)

Figure 9

Scale: 10 mm = 200 N

The compression member = 2000 N

The tension member = 1733 N


13

Equilibrium: task 1 1. Study the following statements and cross out the incorrect answer.

A body that is accelerating is in a state of equilibrium. TRUE/FALSE

For a body to be in a state of equilibrium it is necessary only for the vector

sum of the forces acting on it to be zero. TRUE/FALSE

A resultant force is a single force that can replace two or more forces.

TRUE/FALSE

If two or more forces are replaced by a resultant force, the effect on the body

is changed. TRUE/FALSE

An equilibrant force is the force that, if applied to a body, will cause the body

to be in a state of equilibrium. TRUE/FALSE

The equilibrant force is identical to the resultant force. TRUE/FALSE

2. Try to explain two conditions necessary for a structure or body to be in

equilibrium.

3. Two forces are acting on a body as shown.

(a) Graphically indicate their size and direction.

(b) Graphically indicate the resultant of the two forces.

4. Two forces are acting on a body as shown.

(a) Graphically indicate their size and direction.

(b) Calculate the resultant and direction of the two forces.

5. What are the resultant and equilibrant of the two forces affecting the system

below?


6. A small crane is used on a fishing trawler to lift cases of fish to the dock. The

weight of the lift is 1200 N. Determine the size and direction of the forces in each

of the crane members. (Use a scale where 10 mm represents 200 N.)

7. A weight of 2000 N is suspended by a rope attached to a hook firmly fixed to a

roof joist. A second rope is attached to the vertical rope and pulled horizontally

until the rope makes an angle of 30 to the vertical as shown. Determine the

horizontal pull on the rope and the force on the hook.

8. The figure below shows a cranked lever that is part of a gear-change mechanism.

Find the resultant force FR acting on the hinge pivot and the angle .

FR

600 N

750 N

2000 N

1200 N


15

EFFORT

LOAD

DISTANCEMOVED

BY LOAD

DISTANCEMOVED

BY EFFORT

Levers

Figure 1(a) shows an early lever. The large boulder is too heavy to move by pushing

it. By using a small boulder as a pivot point and a branch as a lever, it is possible to

amplify the force applied to the large rock. The further from the pivot the effort is

applied, the easier it is to move the large rock or load.

Figure 1(a) Figure 1(b)

When a weight is attached to one side of a lever to assist the user, it is known as a

counterbalance.

A universal systems diagram of a lever is shown in figure 2. A lever system changes

an input force and an input motion into an output force and an output motion.

Figure 2

The point that a lever pivots about is called a fulcrum. A line diagram of a lever is

shown in figure 3. The input force is called the effort and the input motion is the

distance moved by the effort force. The output force is called the load and the output

motion is the distance moved by the load.

Figure 3

The lever is a force multiplier and is normally used to get a large output force from a

LEVER SYSTEM

INPUT FORCE

INPUT MOTION

OUTPUT FORCE

OUTPUT MOTION


small input force. However, it can also be used as a distance multiplier, giving a large

output movement for a small input motion; but it cannot do both at the same time.

Figure 4 below shows a lever system designed to move heavy machine castings from

a lower level to a position of installation.

The castings must be lifted 200 mm.

Figure 4: machine-loading lever system

Force multiplier ratio

In the lever system shown in figure 4 above, the load being lifted is about three times

more than the effort being applied. The load divided by the effort gives a ratio. This

ratio is a force multiplier, or how much more load can be lifted compared to the effort.

The lever in figure 4 therefore has a force-multiplier ratio of 2.88 (a ratio has no units

of value).

Example 1 Find the force-multiplier ratio for the lever in figure 4 above.

Force-multiplier ratio = load

effort

= 750 N

260 N

= 2.88

Movement-multiplier ratio

The force multiplier ratio appears to give the user something for nothing. The user is

only applying about a third of the force to move the load. However, it can be seen

from figure 4 that the effort side of the lever has to move much further than the load

side. The ratio of the distance moved by the effort, divided by the distance moved by

the load, is known as the distance-multiplier ratio.

The lever in figure 4 therefore has a distance-multiplier ratio of three (again a ratio

has no units of value).

EFFORT = 260 N

LOAD = 750 N

600 mm


17

Example 2

Find the distance-multiplier ratio for the lever in figure 4 above.

Movement-multiplier ratio = distance moved by the effort

distance moved by the load

= 600 mm

200 mm

= 3

Efficiency

Owing to the effects of friction and inertia associated with the movement of any

object, some of the input energy to a machine is converted into heat, and losses occur.

Since losses occur, the energy output of a machine is less than the energy input; thus

the mechanical efficiency of any machine cannot reach 100 per cent.

The efficiency of a lever system is found by dividing the force ratio by the movement

ratio, with the efficiency given as a percentage. The result of the above division is

multiplied by 100 to give the percentage efficiency value.

Example 1 Find the efficiency of the lever system shown in figure 4.

Efficiency (η) = Force Ratio 100

Movement Ratio

η = 2.88 100

3

η = 96

The system shown in figure 4 has an efficiency of nearly 100 per cent. No system can

be 100 per cent efficient; there are always losses. The losses in a lever system consist

of energy lost to friction at the fulcrum of the lever and the energy lost in strain as the

lever bends slightly. In some cases a small amount of energy will also be lost in the

form of sound.

Remember, no machine is 100 per cent efficient. Common energy losses include heat

energy due to friction, strain energy and sound energy.


Levers: task 1

Draw a universal system diagram of a lever system. Label the diagram input, process

and output.

Complete the line diagram of a lever shown below. You should identify the load,

effort and fulcrum.

INPUT OUTPUT

Levers: task 2 Calculate the force-multiplier ratio of the following levers. Show all working.

EFFORT1OO N

EFFORT300 N

EFFORT200 N

LOAD400 N

EFFORT50 N

LOAD100 N


19

Levers: task 3 A diagram of a lever system is shown below.

(a) Find the force-multiplier ratio of the lever system.

(b) Calculate the movement-multiplier ratio of the lever.

(c) Calculate the efficiency of the system.

(d) Identify possible efficiency losses in the system.

Show all calculations.

(a) Force ratio =

(b) Movement ratio =

(c) Efficiency (η) =

(d) Possible efficiency losses in a lever system =

EFFORT = 150 N

LOAD = 450 N

650 mm

200 mm


Classes of levers

Levers can be divided into three distinct types (classes) determined by the position of

the load, effort and fulcrum. Applications of their use are found almost everywhere,

from the home or school to equipment on the space shuttle. The classes of levers are

as follows.

Class 1 In class 1 levers the effort is on one side of the fulcrum and the load is on the opposite

side (figure 5). Class 1 levers are the simplest to understand: the longer the crowbar

the easier it is to prise open the lid.

EFFORT

LOAD

FULCRUM OR PIVOT

Figure 5

Class 2 In class 2 levers the fulcrum is at one end of the lever and the load and the effort are

spaced out on the other end of the bar. The load must be closer to the fulcrum than the

effort (figure 6). A wheelbarrow is a good example of a class 2 lever. The wheel is the

fulcrum, the load is in the container area and the effort is applied to the handles.

Similarly, a door has a hinge (fulcrum), the load can be considered as acting in the

door’s centre of gravity and the effort is applied as far from the hinge as possible.

Figure 6

Class 3 Class 3 levers are similar to class 2 levers except that now the effort is closer to the

fulcrum than the load (figure 7). This means that more effort has to be applied to

move the load. This type of lever is used when mechanisms require a large output

movement for a small input movement.

EFFORT

LOAD

FULCRUM


21

EFFORT

LOAD

FULCRUM

Figure 7

Examples of various types of lever are shown below; in some cases it is difficult to

tell exactly into which class they fit.

A

B

C

L

F

E

L

F

E

EF

L

E

E

F

L

L

F

E

E

L

L


Levers: task 4

Complete the following list in a table format.

Name the equipment.

Draw a line diagram with arrows showing the fulcrum, effort and load.

Name the class of lever.


23

Levers: task 5

Calculate the force multiplier ratios for the following levers and state which class of

lever each one belongs to.

Class of lever _________

1..

10N

Class of lever _________

F.M.R.


Levers: task 6 In the printed version of these materials, issued to Scottish schools in August 2001, this page contained

embedded copyright material. For copyright reasons that material has been removed for this website

version. In order to see the completed text for this page Scottish schools are advised to refer to their

copy of the printed version. For other users the complete pack is available from Learning and Teaching

Scotland, priced £24.00.

Using your CD-ROM The New Way Things Work, try to answer the following

questions from Principles of Science (Levers) in an interactive way.

1. When rotating a lever mechanism what other name can be used when effort is

applied?

2. Name two related machines that use levers.

3. Explain in your own words how a lever mechanism is used in a car’s clutch

system.

4. What class of lever is a can opener? Sketch a line diagram to show effort, fulcrum

and load.

5. How does a lever system work in a fire extinguisher?

6. What class of lever is used in an aneroid barometer?


25

Moments

The sketch in figure 8 shows a weight attached to a metal rod, and the rod is free to

rotate around a hinge (pivot) P. If the rope holding the weight stationary is cut, what

happens to the rod? If the rope is cut, the force on the weight causes the rod to swing

or turn around the pivot. This ‘turning effect’ is called a moment.

The weight in figure 8 shows a moment of 20 Nm (10 N 2 m). A moment is

measured in newton-metres.

As long as the rope is not cut, the weight and rod are held in balance by the force in

the tie rope.

When any system is in a steady state it is said to be in equilibrium.

Figure 8

The lever system in figure 9 shows a lever that is in a state of equilibrium (balance).

The input force is tending to turn the lever anticlockwise; the load is tending to turn

the lever clockwise. The forces on each end of the lever are exerting a moment: one

clockwise, the other anticlockwise. If the beam (lever) is in equilibrium, both of these

moments must be equal.

Figure 9

FORCE (10 N)

TURNING EFFECT

2 m

HINGE P

ROPE

WEIGHT


The principle of moments states that the sum of the moments must equal zero or the

sum of the clockwise moments must equal the sum of the anticlockwise moments.

The Greek letter stands for ‘the sum of’ and can be used as a shorthand way of

writing the principle of moments:

CWM = ACWM

F1¹ d1 = F2 d2

The force times the distance turning the lever clockwise is equal to the force times the

distance turning the lever anticlockwise. As stated, moments are measured in newton-

metres. It can be seen that the moment on one side of the lever is equal to the moment

on the other side. (Work done = force distance in the direction of motion.)

Example 4 Using the lever system in figure 10, use the principle of moments to show that the

lever is in equilibrium.

Figure 10

Answer

For equilibrium, the CWM = ACWM. A moment is a force multiplied by a

distance

CWM = ACWM

F1¹ d1 = F2 d2

The load is exerting a clockwise moment; that is, it is tending to make the lever turn

clockwise.

Clockwise moment = 200 N 2 m = 400 Nm

The effort is exerting an anticlockwise moment.

Anticlockwise moment = 400 N 1 m = 400 Nm

CWM = ACWM

Therefore the lever is in a state of equilibrium.


27

Moments: task 1 Use a balanced lever similar to the one in the line diagram shown below, which is

available from your teacher. Use a set of weights as a load. Use the spring balance to

apply the effort to the system for each of the load positions.

EFFORT

LOAD

12

3

PIVOT

Suspend the load from position 1 on the load side of the lever. Measure the effort

required to balance the lever using a spring balance. Record the effort in the table

below. Move the load to positions 2 and 3 and record the effort required for balance

each time.

Position 1 Position 2 Position 3

Load

Effort

Calculate the force multiplier ratio for each position.

Position 1

Position 2

Position 3

Complete this statement. As the load gets further away from the fulcrum, the effort

required to balance it …


Example 5 A car footbrake uses a lever action to amplify the force transmitted by the driver to

the braking system when the driver’s foot presses the foot-pedal. If the driver’s foot

can exert a force of 5000 N, what force will be transmitted to the braking system?

100 mm

500 mm

5000 N INPUT

FULCRUM

FORCE TO BRAKINGSYSTEM (LOAD)

Figure 11

This is a class 2 lever. Take moments about the fulcrum to find the force on the

braking system. Notice the distance from the fulcrum to the input is 600 mm.

The input tends to make the lever turn clockwise; the braking system is opposing the

input and so acts to turn the lever anticlockwise.

The principle of moments states that

CWM = ACWM

F1 d1 = F2 d2

5000 N 0.6 m = braking force 0.1 m

braking force = 5000 N 0.6 m

0.1 m

braking force = 30,000 N or 30 kN


29

Moments: task 2

Use the principle of moments to find the missing force or distance in the following

problems. Show all working.

d = ?

d = ?

E = 400 N L = 1200 N

FULCRUM

CWM = ACWM

CWM = ACWM

CWM = ACWM

CWM = ACWM

E = ? L = 5 kN

FULCRUM

0.9 m 0.3 m

E = 50 N L = ?

FULCRUM

200 mm 40 mm

E = 480 N L = 960 N

FULCRUM

300 mm

0.2 m


Moments: task 3 The hand-cutters shown are used to cut thin metal with the effort and load shown.

(a) Draw a suitable line diagram.

(b) What effort will have to be applied if the force required in the hand-cutters to

shear metal is 1.5 kN?

Moments: task 4 The diagram below shows a tower crane carrying a load of 90,000 N. At the other end

a counterbalance load is applied.

(a) Explain why the crane would be unstable without the counterbalance.

(b) Is it an advantage for the counterbalance to be able to move, either towards the

centre of the crane or away from its centre?

(c) The crane in the diagram is lifting a load of 90,000 N, which is 6.3 m away from

the tower. How far from the tower should a 100,000 N counterbalance be placed

so that the crane remains stable?


31

Moments: task 5 A single-lever monobloc tap is shown below.

(a) If the length of the handle is 250 mm and the effort to turn it is 15 N, what

moment would close the tap valve?

(b) What is the benefit of this type of tap?

(c) Where would this type of tap be very useful?

Moments: task 6 When a fish has been hooked, the pull from the fish is 22 newtons at right angles to

the fishing rod. The pivot is at the end of the rod, which is 2.4 metres long. The angler

applies an effort at 0.4 metres from the end of the rod.

(a) Draw a line diagram with dimensions, loads, pivots, etc.

(b) Calculate the anticlockwise turning moment applied by the fish.


(c) Calculate the effort the angler must apply to stop the rod from turning

anticlockwise.

(d) The angler has to exert a greater effort than the load applied by the fish to

maintain equilibrium. Is this an advantage or disadvantage to the angler?

The bell-crank lever

The bell-crank lever shown in figure 12 is used to transmit the input force and motion

through a right angle. It gets its name from part of the bell mechanism used to

summon servants in Victorian houses. By varying the lengths of the two arms of the

bell crank, it is possible to use it either to magnify an input force or to magnify an

input motion.

Figure 12

Example 6 Use the principle of moments to determine the length of the output side of the bell-

crank lever in figure 12.

Calculate the force-multiplier ratio of the lever.

Answer This is a class 1 lever with a right-angled bend. To find the distance ‘d’, take moments

about the fulcrum. Assume the lever is in equilibrium so that the output force opposes

the input force.

CWM = ACWM

F1 d1 = F2 d2

600 N d = 400 N 0.15 m


33

d = 400 N 0.15 m

600 N

d = 0.1 m

The force-multiplier ratio = load

effort

= 600 N

400 N

= 1.5


Linkages

Levers are often linked together to transmit force or motion. A linkage consists of two

or more levers connected together. Linkages are useful for changing the direction of

an input or for giving greater force or distance amplifications.

Five common linkages found in many machines are shown below.

Figure 1 Figure 2 Reverse motion output; Reverse motion output,

distance from fulcrum but fulcrum is nearer

is the same, therefore, the output so the

same force. force is amplified.

Figure 3

Input and output motion are the same, but there is a large amplification of force.

Figure 4 Figure 5

Reciprocating motion transformed Lazy tongs linkage

to rotary motion. for extra reach.


35

Linkages: task 1

A system diagram of a lever mechanism is shown below. The requirements state that

when the lever is pushed down, the output should rise.

The force-multiplier ratio should be 2:1.

MECHANISM

OUTPUT

OUTPUT

INPUT

INPUT

SYSTEM DIAGRAM

LEVER MECHANISMUPWARD

FORCE

DOWNWARD

FORCE

Design a suitable linked lever system that will achieve the desired output.

A sketched diagram should show:

a line diagram

the load, effort and fulcrum

your notional load and effort indicated in newtons

your calculation showing the force ratio.

Answer Evaluate your solution by stating whether the solution satisfied the requirements

identified from the specification in the question.


Linkages: task 2 A counterbalance lever is required for a small city-centre car park.

Requirements The car park is a small, one-person business. The operator wants a system that can be

opened easily by one person. The system must stay open until closed by the operator

and the system must be safe for the operator’s and customers’ cars.

The following criteria have been identified from the requirements.

The car park barrier must:

lift with a small effort

be operated initially by hand

be able to be locked in an upright position

be improved to operate with a simple electronic circuit.

Design a suitable system to satisfy the design requirements. State how you would test

for the following features.

A suitable lever system

Manual lift and lower, showing the force-multiplier ratio

Lock in raised position

Electronic circuit to automate the system

Safety

Draw a sketch of your solution and state whether your system satisfied the criteria.

Write down how you think you could have improved your solution.


37

Free-body diagrams

It is important to isolate different parts of a structure or body from its adjacent

surroundings. In a line diagram this can be done by drawing a free-body diagram,

which is a diagrammatic representation of all or part of the structure showing the

forces affecting it.

Example

Figure 1

If all the visual components acting on the structure or body were removed and

replaced with their force value, a simplified diagram would allow a better

understanding of how the forces are affecting the structure.

Figure 2: free-body diagram

Figure 2 is a simplified free-body diagram of figure 1. The forces representing the bus

and the weight of the bridge act downwards and are taken through the centre of the

bus and the middle of the bridge. Because of the point of contact of the bus, the arrow

is drawn through its centre. The forces Fh and Fv represent the forces that the supports

have on the structure, therefore they also have to be shown. We could be more

detailed and draw the angled support for the bridge in the rock face.

Free-body diagrams: task 1

Draw a free-body diagram of the aircraft indicating the downward forces and

reactions with arrows. Use suitable lettering.

Fv

FbusFbridge

Fh


Beams

Apart from levers, structural beams and beam-related objects are also affected by

forces and turning moments. For a horizontal structure to be stable (in equilibrium)

when being affected by forces in a vertical plane, the following conditions must be

satisfied.

i. The sum of the forces acting upwards must equal the sum of the forces acting

downwards.

upwards forces = downwards forces

ii. The sum of the clockwise moments about any point must equal the sum of the

anticlockwise moments about the same point. That is


(principle of moments)

Beam reactions

Beams, however, have to be supported differently from lever applications and this

determines beam-support reactions. Beams, therefore, are supported in a number of

ways, such as:

(a) simply supported at both ends

(b) built in at both ends (this type of end-fixed beam is called an encastre)

(c) built in at one end only (this type of beam is called a cantilever)

(d) built in at one end only and simply supported at the other.

Examples of these methods are shown below.

Types of beam support

(a) (b)

(c) (d)

At the points of support, the downwards forces acting on the beam are resisted by the

forces acting upwards. These upward forces are known as beam reactions, or simply

the reactions.

M M

M

M


39

Example 1

Determine the reactions R1 and R2 for the simply supported beam (beam weights will

be ignored in this case).

Figure 1: free-body diagram

Take moments about R1


(10,000 N 2 m) + (500 N 2.5 m) + (6000 N 4 m) = R2 5 m

R2 = 20,000 Nm + 1250 Nm + 24,000 Nm

5 m

= 9050 N

Also upwards forces = downwards forces

R1 + 9050 N = 10,000 N + 500 N + 6000 N

R1 = 16,500 N – 9050 N

= 7450 N

Therefore the reactions for the beam supports are R1 = 7450 N and R2 = 9050 N


Beams: task 1 1. The span of a cantilever diving board is two metres and the downward load of the

diver is 800 N.

(a) What is the maximum reaction force in the board?

(b) Draw a suitable free-body diagram.

(c) What is the minimum reaction, R1, at the fixed end?

(d) Why would this not be a fixed reaction?

2. A beam is simply supported at each end with a span of three metres. The beam

carries a small lifting device having a weight of 1 kN.

(a) Complete a suitable free-body diagram.

(b) When the lifting device is positioned at the mid-point of the beam and carries a

casting weighing 2.5 kN, what are the reactions at R1 and R2?

(c) When the lifting device is positioned one metre from one end and carries a

machine component weighing 6 kN, what are the reactions at R1 and R2?

3. The figure below shows a clamp on a milling machine table for fixing a

component for machining. A clamping force of 1200 N is applied by the bolt to

the component and rear-distance piece when the nut is fully tightened.

100 mm 75 mm

A C

COMPONENT


41

(a) Draw a free-body diagram to show the arrangement of the forces.

(b) Calculate the clamping forces on the component (RA) and the distance piece

(RC).

(c) How could the arrangement be altered to give a bigger clamping force on the

component?

4. The diagram below shows a free-body diagram of a horizontal beam, seven metres

long, which is part of a bridge structure. The beam is simply supported at A and

D. Determine the reaction forces at A and D.

5. The supermarket trolley shown is a form of cantilever.

(a) Sketch the free-body diagram to indicate the major forces.

(b) If the groceries are spread throughout the trolley, can it tip over? If not, why?

(c) What happens if someone leans on the back of the trolley?

(d) What happens if someone applies weight to the front of the trolley?

Refer to your free-body diagram in your answers.


6. The forklift truck must have a minimum downward force of 800 N acting through

the rear wheels.

(a) Draw an appropriate free-body diagram.

(b) Calculate the weight required to balance the load on the lift with R2 = 0 N.

(c) Find the additional weight acting through the centre of gravity of the truck to

produce 800 N at the rear wheels.

7. The total downward load when the truck shown below is empty is 30 kN and

when fully loaded 55 kN. Draw a suitable free-body diagram. Find the reaction in

each of the axles when the truck is empty and when fully loaded.


43

8. A car has been raised on a ramp to look at the drive shaft. The downward load on

the car’s rear and front axles are 5970 N and 3980 N, respectively. The wheelbase

of the car measures 2.5 m.

(a) Draw a free-body diagram.

(b) What is the reaction at R1?

(c) What distance (x) will R1 have to be from the front axle to maintain

equilibrium?

9. The car and caravan shown below have a ball-jointed tow-bar that connects the

car and its caravan. The weights of each are shown, together with the reaction

forces in the centre of all three wheels.

(a) Draw a free-body diagram for the car and caravan.

(b) Looking at the caravan, calculate the force acting at the tow-ball.

(c) Calculate the reaction forces R1 and R2.


Gears

Gears are toothed wheels designed to transmit rotary motion and power from one part

of a mechanism to another. They are fitted to shafts with special devices called keys

(or splines, etc.) that ensure that the gear and the shaft rotate together. Gears are used

to increase or decrease the output speed of a mechanism and can also be used to

change the direction of motion of the output.

The type of gear wheel most commonly used is the spur gear.

Simple gear train

Gears work by interlocking or ‘meshing’ the teeth of the gears together as shown in

figure 1.

Figure 1

When two or more gears are meshed they form a ‘gear train’. The input gear, which

causes the system to move, is called the ‘driver’; the output gear is called the ‘driven’.

Both gears are mounted and supported on separate shafts.

Example

Figure 2 below shows a method of increasing the output speed of a mechanism.

Figure 2

If driver gear A has 24 teeth and it makes one complete turn, then 24 teeth will have

passed point X on the diagram. If driven gear B is meshed with driver gear A, then for

every tooth of gear A to pass point X, one tooth of gear B must pass this point.


45

If 24 teeth of gear A pass point X, then 24 teeth of gear B must pass point X. To be

able to do this, gear B must make two complete revolutions but in the opposite

direction.

Movement-multiplier ratio in gears

The ratio of change in speed between the gears is called the movement-multiplier

ratio.

The ratio of a gear system is found by dividing the number of teeth on the driven gear

by the number of teeth on the driver gear. This can be used to calculate the output

speed of a gear system.

Movement ratio = number of teeth on driven gear

number of teeth on driver gear

Example

For the gear system shown in figure 2 the gear multiplier ratio is

1:2or2

1

24

12RatioGear

This means that if gear A was rotating at 100 rpm (revolutions per minute) clockwise

then gear B would rotate at 200 rpm anticlockwise.

Gears can also be used to decrease the speed of a mechanism, as shown in figure 3.

Figure 3

2:1or1

2

12

24RatioGear

If gear A is still rotating at 100 rpm in a clockwise direction then gear B will now

rotate at 50 rpm in an anticlockwise direction. It is sometimes necessary to obtain a

change in speed without changing the direction of the driven gear. How can this be

done?


Idler gears

To get the driven gear to rotate in the same direction as the driver, a third gear is

inserted in the system. This idler gear has no effect on the multiplier ratio of the

system. The size of the idler is not important and is normally a small gear, as in

figure 4.

Figure 4

The multiplier ratio for the simple gear train in figure 4 is still 2:1. If gear A still

rotates at 100 rpm clockwise then the output of gear B will rotate at 50 rpm

clockwise.

Ratchet and pawl

A wheel with saw-shaped teeth round its rim is called a ratchet. The ratchet wheel

usually engages with a tooth-shaped lever called a pawl. The purpose of the pawl is to

allow rotation in one direction only and prevent rotation in the opposite direction. A

ratchet and pawl mechanism is shown in figure 5.

Figure 5


47

A crane-winding mechanism shown in figure 6 makes use of a ratchet and pawl to

allow rotary motion in one direction only. The crane can be wound up, but the tension

force in the cable cannot unwind the winch because of the ratchet mechanism.

Figure 6

Task 1: simple gear train Using the mechanical components within Crocodile Clips build a simple gear train,

similar to the ones in figure 7, where the driven gear will rotate at twice the speed of

the driver gear. (Use the 1 rpm motor.)

Figure 7

Gears: task 2 You know how to build a simple gear train that will increase the speed of rotation of

the driven gear compared to the driver gear.

From a selection of four gear wheels – 8 t, 16 t, 24 t and 40 t – design and build a

simple gear train that will provide the biggest increase in speed between the driver

and driven gears.

Sketch your results and calculate the multiplier ratio of your system.

(A circle can represent a gear wheel.)

Gears: task 3

Modify your simple gear train so that it will give you the biggest decrease in speed

between the driver and driven gears, but this time with both the input and output gear

rotating in the same direction.

Sketch your results and calculate the multiplier ratio of your system.

CABLE

RATCHET

PAWL BAR

CRANK HANDLE

WINCH DRUM


Gears: task 4

In the printed version of these materials, issued to Scottish schools in August 2001, this page contained






questions from Principles of Science (Gears and Belts) in an interactive way.

1. What are the four types of gear systems listed?

(a) Explain how two of them operate.

2. What is the gear ratio when two 30-teeth bevel gears come into contact?

(a) What action do bevel gears carry out?

(b) What is another name for bevel gears?

3. Does the spur gear on a lawnmower rotate quicker or slower than the larger-

geared roller wheel?

4. Explain how a worm and worm wheel (crank) are used in a windscreen wiper?

(a) What is the gear ratio if the worm wheel has 17 teeth?

(b) How could the wiper be made to work at different speeds?

5. What is a differential on a car?

(a) What types of gears are used?

6. Sketch the graphical symbol when gears determine the movement of clock hands

from seconds to minutes in a mechanical clock.

(a) Show how the ratios of the system can be calculated.


49

AB

C

AB

C

D

+ + + +

Gears: task 5

Calculate the multiplier ratio for the simple gear train below and then find the output

speed and direction if gear A rotates at 250 rpm in a clockwise direction. Show all

your working.

A = 20 teeth

B = 5 teeth

C = 30 teeth

Answer Multiplier ratio

Output speed

Gears: task 6 For the simple gear train shown below, find the following.

(a) The gear that rotates in the same direction as A.

(b) The multiplier ratios of A to B, A to C and A to D.

(c) The speed of B, C and

D if A rotates at

500 rpm.

A = 50 teeth

B = 10 teeth

C = 25 teeth

D = 100 teeth

Answers

(a)

(b) A to B A to C A to D

(c) B = C = D =


Compound gears

If gears are required to produce a very large change in speed, for example if the

multiplier ratio is 100:1, then problems can arise with the size of gear wheels if a

simple gear train is used. This problem can be overcome by mounting pairs of gears

on the same shaft, as shown in figure 7.

This arrangement is described as a compound gear train. This type of gear train can

also be used to provide different outputs moving at different speeds and in different

directions.

Figure 7

The compound gear system in figure 8 shows how the shafts are connected between

the ‘pairs’ of gears. Gears B and C are connected and rotate at the same speed. To

calculate the multiplier ratio for the gear train it is necessary to calculate the ratio for

each pair of meshing gears.

Figure 8

Example

The multiplier ratio for the system shown in figure 7 is as follows.

The multiplier ratio for the first pair of meshing teeth is

1:420

80

driver

drivenAB of Ratio

The multiplier ratio for the second pair of meshing teeth is

1:610

60

driver

drivenCD of Ratio


51

The total multiplier ratio is calculated by multiplying both ratios:

1:241

6

1

4ratioTotal

For an input speed of 100 rpm, the output speed would be less than 5 rpm, that is,

4.17 rpm.

Worm and wheel

Another way of making large speed reductions is to use a worm gear and wormwheel,

as shown in figure 9. The worm, which looks rather like a screw thread, is fixed to the

driver shaft. It meshes with a wormwheel, which is fixed to the driven shaft. The

driven shaft runs at 90 degrees to the driver shaft. When considering the speed

changes in most worm gear systems, you can think of the worm as if it were a spur

gear with one tooth. It is a single tooth wrapped around a cylinder.

Figure 9

Example

The multiplier ratio between the gears in figure 9 is

1:301

30

driver

drivenratioMultiplier

This would mean that for a motor rotating at 100 rpm, the output driven gear would

rotate at only 3.33 rpm.


Bevel gears

Bevel gears, like worm gears, use shafts at 90 degrees to each other, as shown in

figure 10.

Figure 10

The food whisk shown in figure 11 uses bevel gears not only

to change rotary motion through 90 degrees, but also, by using

different sized gears, to increase the speed of rotation.

The one shown gives a speed increase of 1:5. Figure 11

Gears: task 7 Produce the greatest possible speed within a compound gear train using spur gears

with 8 t, 16 t, 24 t and 40 t. This can be done using computer simulation if available

with the 1 rev motor constant speed motor as a power source.

Complete the following.

Sketch or print out your results.

Sketch your gear train graphically (as in figure 8).

Calculate the multiplier ratio for your system.

Gears: task 8

Two pairs of bevel gears, all of equal size, are used to model the wind generating

system shown below. The output from these bevel gears can be connected to the

compound gear system of the previous assignment. Calculate the output speed if the

vanes of the windmill are rotating at 10 rpm.


53

MOTOR

A B

CD

OUTPUT

GENERATOR

Gears: task 9 The compound gear train shown below is driven by a motor that runs at 1000 rpm.

Calculate the multiplier ratio of the motor to the output shaft and then the output

speed. Show all your working.

A = 20 teeth

B = 60 teeth

C = 40 teeth

D = 50 teeth

Answer

Multiplier ratio =

Output speed =

Gears: task 10 A motor with a single worm wheel output rotates at 500 rpm. You are given the

following sizes of worm gears from which to select.

(a) = 10 teeth

(b) = 25 teeth

(c) = 50 teeth

Explain which gear should be connected to the motor to give the slowest output speed

and why. What is the output speed?


A

MOTOR

B

C

D

E

F

DRUM

LOAD

Gears: task 11

The motorised winch shown below runs at a speed of 1200 rpm. The drum is to rotate

at 25 rpm. Calculate:

(a) the multiplier ratio required to produce the speed reduction

(b) the number of teeth gear A must have to meet this requirement.

A = ?

B = 32 teeth

C = 15 teeth

D = 45 teeth

E = 12 teeth

F = 48 teeth

Answer

Movement ratio =

Number of teeth in A =

Also calculate for the above system the following.

If the radius of the drum is 50 mm, what is the speed of the load being raised?

(Answer in m/s)

Answer

Lifting speed =


55

Gears: task 12 In the printed version of these materials, issued to Scottish schools in August 2001, this page contained






questions from Principles of Science (Gears and Belts) in an interactive way..

1. How is a ratchet and pawl used in a car sear belt?

2. Explain how Derailleur gears and the shifter change the gears in a bicycle.

3. What selection of gears should be made to cycle:

(a) on a level surface?

(b) uphill?

(c) downhill?

4. What type of mechanism could be used in a window roller blind? How does the

roller blind work?

5. A simple mechanism is used to assist a car’s steering.

(a) What is this mechanism?

(b) How does it work?

(c) What other mechanical function is used in the steering arm?

6. Transferring forces over distances is achieved in wind turbines and windmills.

What mechanical device is used and why are they selected?

7. A belt driven fan is used in a car’s cooling system. Why?


8. A stepper motor is used in a computer system’s disc drive.

(a) What mechanical system does it drive?

(b) What effect does it have on the system?

9. An escalator uses two types of mechanical systems. What are they?

10. A belt drive is used from a motorised spindle in a record player turntable. Does

the belt speed up or slow down the turntable?


57

Torque and Drive Systems

Torque is the amount of turning produced by a force. The turning or twisting action

exerted by a force or number of forces will cause or tend to cause rotary motion.

Drive shafts in cars, tools turning, belt-and-pulley systems, etc. are all affected by

torque.

A simple example of this is when the propeller of a model builder’s toy boat connected

to a rubber band is twisted by torsion forces. When the propeller is released, the rubber

band, having been under the twisting effect, releases energy to drive the boat through

the water.

Example 1 How much torque is required to tighten the nut if the force required is 45 N and the

radius of the tool is 200 mm.

Figure 1

Torque = force radius

= 45 N 200 mm

Example 2 A belt drives a pulley with a diameter of 200 mm. If the effective belt tension tending

to turn the pulley is 250 N, find the work done per revolution.

When a force of P newtons acts at the rim of a pulley of r metres radius, then the work

done per revolution is r2P ; that is, P newtons circumference (2r).

Therefore, the work done per revolution

= torque (Pr) 2

J157

m1.0N25014.32


Power transmitted by a belt drive

Example 3 The effective pull on a belt drive is 420 N when driving a 500 mm diameter pulley.

The speed of rotation is 220 revolutions per minute. Find the power.

When a force, P newtons, acts at the rim of a pulley, of r metres radius, revolving at n

revolutions per second, the power or work done per second is given by nr2P .

Power =force (P newtons) circumference (2r) revolutions/s (n)

Thus power, or work done/s = torque (Pr) angle rotated through/s (2n)

= 2nT

The effective driving torque = force radius

= (T1 – T2) diameter (d)

2

T1 is the tension on the tight side.

T2 is the tension on the slack side.

Therefore power transmitted = dn (T1 – T2)

kW42.2orwatts2140

42060

2205.014.3

TTndPower 21

Torque: task 1 (a) Calculate the power transferred if a 230 mm diameter pulley wheel revolves at

25 revolutions per second. The pulley has one belt and the tension in the tight side

of the belt is 436 N, while in the slack side it is 186 N.

(b) A shaft transmits 18 kW when rotating at 200 rpm. What is the torque in the shaft?

(c) A railway traction motor develops 150 kW when the train moves along the track.

The rail wheel rotates at 1500 rpm. Find the torque in the driving axle.

(d) An electric motor exerts a torque of 23 Nm and rotates at 2800 rpm. Find the

power of the motor.

(e) The effective pull on a belt is 360 N when driving a 400 mm diameter pulley. The

speed of rotation is 250 rpm. Calculate:

the power without slip

the power with three per cent slip.

(f) During a machining test on a lathe, the tangential force on the cutting tool was

found to be 220 N. If the work-piece diameter was 120 mm, what was the torque

on the lathe spindle?


59

Belt-and-chain drives

Many mechanisms make use of rotary motion, often provided by someone turning a

handle or by an electric motor. But to be useful, this rotary motion has to be frequently

transmitted from one part of a mechanism to another, often with a change of speed.

While gears can be connected together in a simple gear train, if too many gears are

used there can be large efficiency losses due to friction.

There are two simple means of transmitting rotary motion over relatively large

distances. One is to use a belt wrapped around two or more pulleys as shown in figure

1. The belt is tightened or tensioned by pulling one of the pulleys out and locking it in

place. Pulleys are thin metal discs with a groove cut into the circumference of the disc.

Figure 1: belt-and-pulley symbol

The tensioned belt transmits the rotary motion from pulley 2 to pulley 1. The belt is

angled as shown in figure 2 to give better grip to prevent the belt from slipping. A

change in speed can be accomplished by varying the diameter of the driver pulley and

driven pulley.

Figure 2: vee belt for extra grip

Changes in direction can be achieved by crossing the belt as shown in figure 3.

In belt-drive systems, the belt must be crossed between the two pulleys if the direction

of the output shaft is to be opposite to that of the input shaft.

40 mm

160 mm

DRIVEN PULLEY

DRIVER PULLEY

1

2


DRIVEN

DRIVER

Figure 3

Belt drives are used in a wide variety of situations. They are made from a composite of

two materials, rubber and string. The string helps to prevent the rubber from stretching

too much. Drive belts are inexpensive to produce. They are easy to replace and need

little maintenance, as they do not require lubrication. They also absorb shock loads.

For instance, if a belt drive is used to transmit the power from a motorcycle engine to

the rear wheel and the biker tries to ‘wheelie’, the belt tends to slip, preventing damage

to the engine. Belt drives are found in many household machines such as washing

machines, vacuum cleaners (figure 4), tumble dryers and so on.

Figure 4: vacuum cleaner drive belt

Drive systems: task 1 Many machines and mechanisms use belts and pulleys to transmit rotary motion. Write

down any machines or mechanisms that you know of which use belts and pulleys.


61

Drive systems: task 2 Draw a universal systems diagram for one of your above answers.

Drive systems: task 3 Draw a symbol for two pulleys that produce a decrease in speed and with a change in

direction for the driven pulley.

Multiplier ratio for belt drives

Pulley systems can be used to transmit rotary motion over a large distance. The input

rotary motion is often from a fixed-speed and fixed-torque electric motor. Torque is a

turning force produced by mechanisms and is measured in newton-metres (Nm).

Changing the ratio of the diameters of the pulleys can vary both the speed of the

output and the torque at the output.

40 mm

120 mm

DRIVEPULLEY

MOTOR

Figure 4: belt-and-pulley system


Example The motor in figure 4 is connected to a pulley of diameter 120 mm. This is the driver

pulley. The driven pulley has a diameter of 40 mm. The multiplier ratio of the pulley

system is the diameter of the driven pulley divided by the diameter of the driver

pulley.

Multiplier ratio = diameter of driven pulley

diameter of driver pulley

For the system in figure 4 the multiplier ratio is 40 = 1 or 1:3

120 3

Example

Motor speeds

If the motor speed is 1200 rpm, the output can be found by dividing the input speed

by the multiplier ratio.

The output speed can also be found from the multiplier ratio: input speed

output speed

Output speed = input speed

multiplier ratio

Output speed = 1200 rpm

1/3

Output speed = 3600 rpm

In figure 4 the speed of the motor is increasing; there must be some loss to compensate

for this gain. The loss is in output torque. In general, as the output speed increases, the

torque decreases. As the speed decreases, the torque increases and this affects the

turning force. Electric motors are rated at certain torques for specific voltage supplies.


63

Drive systems: task 4

Label the line diagram of the belt-drive system shown below using the following

terms.

driver pulley

driven pulley

belt

INPUT OUTPUT


(a) In the above system, when the driver is turned, does the driven pulley turn faster or

slower than the driver?

Answer

(b) If the diameter of the driver pulley is 40 mm and the diameter of the driven pulley

is 10 mm, what is the multiplier ratio?

Answer

(c) If you placed a chalk or tape marker at the top dead centre of each of the two

pulleys and turned the driver pulley once, how many revolutions would the smaller

driven pulley make?

Answer

Example

Figure 5 shows a belt-drive system for transmitting rotary motion from an electric

motor to a spin-dryer system in a washing-machine drum. The motor has an output

torque of 800 Nm at 1000 rpm.

Calculate the multiplier ratio of the system, the speed of the drum and the output

torque produced by the drum.


Ø50 mm

Ø250 mm

ELECTRIC MOTOR1000 RPM

SPIN DRUM

Figure 5: washing-machine spin dryer

Answer

1:5

50

250

pulleydriverofdiameter

pulleydrivenofdiameterratioMultiplier

rpm200

5

rpm1000

ratiomultiplier

speedinputdrumtheofspeedoutputThe

Nm4000

5Nm800

ratiomultipliertorqueinputtorqueOutput

A variety of output speeds and output torques can be achieved by using stepped-cone

pulleys, as shown in figure 6. The drive motor is attached to one set of pulleys and the

drive belt can be moved between the various pairs of pulleys to give a selection of

speeds.

Figure 6: stepped-cone pulley system


65

One of the advantages of belt drives is that they will absorb shock loads by slipping.

However, excessive slipping will create inefficiency in the system. At the same time,

if the belt is too tight the pulley bearings could be damaged. One method of keeping

the belt correctly tensioned is to use a spring-loaded jockey pulley, as shown in figure

7.

JOCKEYPULLEY

DRIVERDRIVEN

Figure 7: a jockey pulley for tensioning

Toothed belts

Belt drives tend to use their ability to slip to their advantage. However, where slippage

would damage a mechanism, toothed belts have been developed that retain the

advantages of normal belts but do not slip.

Many cars have toothed belts (for example timing belts) to control the opening and

closing of the inlet and outlet valves in the car engine. If the belt slipped, the pistons

would collide with the valves, damaging the engine. These belt drives are quiet,

require little maintenance and are easily changed if required (figure 8).

Figure 8: toothed belts

Chain drives

Where large forces have to be transmitted, and there can be no slippage allowed,

chain drives are used. Instead of a pulley, a toothed wheel known as a sprocket is used

to drive a chain. The chain in turn drives another toothed wheel. Once again, the

speed can be varied by making the sprockets different sizes.


Figure 9: Bicycle-chain drive

Figure 9 shows an application of a chain drive that is familiar to everyone. This can

help to illustrate the advantages and disadvantages of chain drives. When cycling, if

you want to go faster suddenly, you stand up and put extra weight (force) into the

pedals. This force is transmitted to the back wheel by means of the chain. If the chain

were to slip, what would happen? Unless the chain and sprockets are worn, the chain

will not slip and the extra force will carry out its task in allowing you to go faster.

Chains are very strong, and unless badly worn, they will not slip. However, they have

to be oiled regularly, and both the chain and sprockets are prone to wear. They are

also more expensive to make and buy than belt drives. Chain drives are also much

noisier that belt drives.

Drive systems: task 6 Look at the chain drive shown below.

DRIVEN

DRIVER

(a) When the driver is turned, does the driven gear turn faster or slower than the

driven sprocket?

(b) If a mark was placed at the top of the large and small sprockets and the driver

sprocket rotated, how many times would the driven sprocket rotate?

(c) Explain in technological language how the chain could be kept at the correct

tension.


67

(d) What is lubrication and why is it important to keep the chain well lubricated?

(e) Draw a system diagram for a tensioned chain drive.

(f) Is the above system an open or closed looped system?

Multiplier ratio for chain drives

Calculating the multiplier ratio, output speed and torque of a chain drive system is

very similar to calculating them in belt-drive systems.

Example A pedal cycle has 60 teeth on the driver sprocket and 10 teeth on the driven sprocket.

What is the multiplier ratio of the chain-drive system?

6:1

60

10

sprocketdriveronteethofnumber

sprocketdrivenonteethofnumberratioMultiplier

Chain tension

Chain-drive systems must also have a means to tension the chain. If the chain is over-

tensioned there will be excessive wear on the chain, sprockets and bearings in the

system. In some bicycles and even motorcycles, the chain is tensioned by gently

pulling the wheel back until the chain is tight and then tightening the locking wheel

nuts. However, to give better control, a spring-loaded jockey wheel such as that used

in Derailleur gears on racing bikes and mountain bikes is used, as shown in figure 10.

Figure 10: Derailleur gears


Example

The bicycle shown in figure 11 has two rear sprockets, one with 50 teeth and the other

with 80 teeth. The driver sprocket has 200 teeth. Calculate the output torque for the

two rear sprockets if the input torque is 20 Nm.

Figure 11: a two-gear bicycle

Answer First find the multiplier ratio for the two driven sprockets.

4:1

200

50


sprocketdrivenonteethofnumbersprocketsmallofratioMultiplier

52:1

200

80


sprocketdrivenonteethofnumbersprocketearglofratioMultiplier

The output torque for each size of sprocket can now be found.

Nm5smalltorqueOutput

4:1Nm20

ratiomultipliertorqueinputsprocket) (small Torque

Nm8eargltorqueOutput

52:1Nm20

ratiomultipliertorqueinputsprocket) (large Torque


69

Example

A motorcycle uses a belt drive to transmit power from the engine to the rear wheel as

shown in figure 12. If the engine rotates at 3000 rpm, what will be the rotary speed of

the rear wheel?

150 mm

50 mmWHEELPULLEY

ENGINE DRIVEPULLEY

Figure 12: motorcycle belt drive

Answer The rotary speed of the driver pulley multiplied by the diameter of the driver pulley is

equal to the rotary speed of the driven pulley multiplied by the diameter of the driven

pulley.

Rotary speed of driver pulley = Rotary speed of driven pulley

diameter of driver pulley diameter of driven pulley

rpm1000

150

503000

pulley driven ofdiameter

pulleydriver ofdia pulleydriver of speedrotary pulley driven of speedRotary

The rotary speed of the rear wheel is 1000 rpm.



Calculate the multiplier ratios for the following belt-drive systems.

They are driven from A to B. Also indicate with an arrow the direction of rotation of

B assuming A is clockwise.

A

B

30 150

A

B

160

100

25

5

A

B100 30 4

80

A

B

30

120

4

40

160

48


71

Converting motion

We know that there are four kinds of motion. These comprise:

(a) rotary

(b) linear

(c) reciprocating

(d) oscillating.

Many mechanisms involve changing one type of motion into another. For example,

the rotary motion of a pillar-drill handle is changed to the linear motion of the chuck

and drill bit moving towards the material being drilled.

Cams

A cam is a specially shaped piece of metal or plastic which can be used to change an

input rotary motion to an output motion that is oscillating or reciprocating.

The cam operates by guiding the motion of a follower held against the cam, either by

its own weight or by a spring. As the cam rotates, the follower moves. The way that it

moves and the distance it moves depend on the cam’s shape and dimensions.

The two main types of cam and follower are shown below.

1. The circular or eccentric cam (figure 1)

2. The pear-shaped cam (figure 2)

ECCENTRICCAM

FOLLOWER

ROTARY

CAMROTARYMOTION

FOLLOWER RECIPROCATINGMOTION

Figure 1 Figure 2

Other, more complex, shapes can also be used.

Cam motion

Pear-shaped cams are often used for controlling valves. For example they are often

used on motor-car camshafts to operate the engine valves. A follower controlled by a

pear-shaped cam remains motionless for about half a revolution. During the time that

the follower is stationary, the cam is in a dwell period. During the other half-

revolution of the cam, the follower rises and then falls. As the pear-shaped cam is

symmetrical, the rising motion is the same as the falling motion.


CAM

ROTATION

VALVE OPEN

STROKE

VALVE CLOSED

Figure 3 Figure 4 Figure 5

Figure 3 shows the valve fully opened as the follower is in contact with the highest

point of the cam, its crown.

Figure 4 shows the valve closed as the follower is in contact with the lowest point of

the cam, its heel.

Figure 5 shows the valve about to open at the end of its dwell period.

When not on the ‘dwell’ part of the cam cycle, the

follower rises and falls and the valve opens and

closes. The distance between the highest and

lowest points on the cam profile is called the

stroke of the cam. The distance the valve opens is

the same as the stroke of the cam.

In a car engine, cams are fixed on a camshaft. As

each cylinder has two valves, an inlet and an

exhaust valve, there are two cams on a camshaft

for each cylinder, as shown in figure 6.

Figure 6


73

Crank slider

Crank slider mechanisms involve changes between rotary and reciprocating motion,

as shown in figure 7. The crank rotates while the slide reciprocates. The longer the

crank the further the slider will move. The two main ways that crank-slider

mechanisms are used are described below.

CRANK

SLIDER

Figure 7

1. Reciprocating motion to rotary motion

Car engines use reciprocating pistons, which are connected to a crankshaft by

connecting rods, as shown in figure 8. As the pistons move up and down the

connecting rods push the crankshaft round. Each piston moves down in turn, so

keeping the crankshaft turning.

Figure 8 Figure 9

2. Rotary motion to reciprocating motion

A power hacksaw, shown with guards removed in figure 9, uses an electric motor

to power a crank, which is connected to a saw frame. The saw frame is free to

slide on the ‘arm’. As the crank rotates it causes the frame to slide backwards and

forwards on the arm. The longer the crank the further the saw frame will move.


Converting motion: task 1

The pear-shaped cam and follower shown represent a simple locking mechanism.

LOCK BOLTA

B

1. Name parts A and B

A

B

2. How much of a turn does the wheel have to make to push the lock-bolt closed?

Tick the correct answer.

¼ turn ½ turn 1 turn 2 turns

3. Complete the system diagram for the movement of the lock.

Input Output

Lock

Motion Motion

4. What does the spring do?


75

Converting motion: Task 2 The cam-and-valve mechanism is part of a car engine and is shown in figure 2.

Complete the systems diagram to show the input and output motion of the mechanism.

CAM

VALVE

SPRING

COLLAR

42 mm

20 mm

Figure 10

_______________motion ________________motion

If the cam on the valve mechanism turns half a revolution from the position shown on

the diagram, what distance does the valve move?

___________________mm


Converting motion: task 3 A crank-and-slider mechanism is used in a fabric-testing machine, as shown in

figure 11.

Figure 11

_______________motion ________________motion

(a) What is the distance from A to B?

(b) What effect does the wire brush have on the fabric?

50 mm


77

Rack and pinion

A rack-and-pinion mechanism is used to transform rotary motion into linear motion,

or linear into rotary motion. A round spur gear, the pinion, meshes with a ‘rack’ that

can be thought of as a spur gear with teeth set in a straight line (figure 1).

PINION

RACK

Figure 1

Gear wheels are normally made from metal or plastic. Plastic gears have the

advantage that they are much quieter running and need less lubrication.

The rack and pinion can transform rotary motion into linear motion and linear motion

into rotary motion in three ways.

1. Movement of the rack in a straight line causes the pinion to rotate about a fixed

centre (figure 1 above).

2. Rotation of the pinion about a fixed centre causes the rack to move in a straight

line as used in a pillar drill (figure 2).

Figure 2


3. If the rack is fixed and the pinion rotates, then the pinion’s centre moves in a

straight line, taking the pinion with it like the movement of the carriage along the

bed of a centre lathe (figure 3).

Figure 3

Rack and pinion: task 1

A rack with 100 teeth per metre is meshed with a pinion that has 10 teeth.

Figure 4

1. If the pinion rotates one revolution, how far does the rack move?

2. How many revolutions does it take to move the rack from one end to the other?

RACK

CARRIAGEBED

HEADSTOCK

PINION BEHINDAPRON


79

Rack and pinion: task 2

A rack with 100 teeth per metre is meshed with a pinion that has 10 teeth.

1. If the pinion rotates one revolution, how far does the rack move?

2. How many revolutions does it take to move the rack from one end to the other?

3. Figure 2 below shows a rack and pinion mechanism being controlled by a stepper

motor. If the movement of the motor is 7.5 degrees per pulse, what is the number

of pulses required to move the rack 50 mm?

Figure 5


Rack and pinion, cams and cranks: task 3

In the printed version of these materials, issued to Scottish schools in August 2001, this page contained





Using your CD-ROM: The New Way Things Work, try to answer the following

questions from Principles of Science (Cams and Cranks) in an interactive way.

1. Look at how cams are operated. For what two operations are cams used?

2. Name the two types of crank input system. Name one type of movement

conversion that takes place.

3. Draw a system diagram for a windscreen wiper. How does the system work in

terms of the crank and the connecting rod?

4. What does a crankshaft do in a four-stroke car engine?

5. Cams are connected to the camshaft in a four-stroke car engine. What is their

purpose?

6. An electric motor drives a crank in an electric shaver. How does this affect the

cutting process?

7. A shutter in a movie camera uses a crank. Explain what it is used for.

8. A gauge, which uses gears and a lever as a crank, is used to read the depth of a

diver. What is it called?

9. Draw a system diagram for a lawn sprinkler. What does the crank mechanism do

in a systems operation?

10. What effect does a cam have on a cylinder-lock door?


81

Couplings

Rotary machines employ a variety of methods of transmitting motion from one part of

a machine to another. The motion is often transmitted through shafts, which are round

metal rods. Often these shafts must be connected together to transmit the motion.

Shafts are joined using a device called a coupling. In small models, such as those used

in schools, simple sleeves or tubes of plastic use friction to drive two shafts, which are

pressed into the sleeve. Stronger couplings are required for industrial-sized machines.

Aligned shafts Where shafts are in line with each other they are joined either with a flanged coupling

or a muff coupling. All couplings must be ‘keyed’ to the shafts they are joining to

give a positive drive. Figure 1 shows a flange coupling and a muff coupling.

Figure 1(a): flange coupling

Figure 1(b): muff coupling

Non-aligned shafts Where shafts meet at a slight angle, some method of compensating for misalignment

must be used. Where the misalignment is small, a flexible coupling (flexi-coupling),

using either rubber or a mixture of rubber and steel, is used. The rubber is flexible

enough to compensate for small changes in angle (figure 2).

SPIDER

FLEXIBLEDISC

Figure 2: flexi-coupling


When the alignment is more than a few degrees out, a universal joint is used. A

universal joint can transmit motion through an angle of 20 degrees. Figure 3 shows

Hooke’s universal joint. The two yokes are free to pivot on the central ‘spider’.

Modern universal joints use needle roller-bearings between the spider and the yokes.

Figure 3: universal joints

Bearings

Parts of mechanisms that slide over each other use flat bearings. Flat bearings tend to

be made from cast iron, brass or bronze. Brass and bronze bearings, which are softer

than the materials sliding through or over them, will wear. They are sometimes called

wear strips. When badly worn they are replaced. Cast iron is a self-lubricating material

and is very strong when compressed.

RAMSLIDINGBEARING

SURFACES

BEARINGADJUSTMENT

MACHINECASTING

Figure 1: a flat bearing and wear strip

When a shaft is turned, it must be supported in some way. Friction opposes motion,

and when a shaft is turning there is likely to be heat and wear at the supports. The

amount of heat and wear due to friction will vary with the materials used, the forces

involved and the speeds involved. Various types of bearing and bearing materials have

been developed to reduce friction in mechanisms.

Bearings that support a round shaft are called journal bearings. When a journal bearing

has to take some axial load, it must have a shoulder to take this load. When a shaft has

a large axial load, it must have a thrust bearing. Figure 2 shows a combined thrust and

journal bearing.


83

Figure 2: combined thrust and journal bearing

Journal bearings are made from a variety of materials: the most common are bronze

and white metal. Bronze is used where slow, heavy loads are carried. White metal, an

alloy of tin, copper and antimony, which is soft and melts when overheated, is used in

systems with light loads. Plastic and nylon bearings are also very common.

Split bearings As bearings are designed to wear, it stands to reason that they must be able to be

removed and replaced. When the bearing support is at the start or end of a shaft, it is

simple to remove and replace it. However, when a shaft is very long, it may be

supported at several points along its length. To make it easy to remove and replace

bearings, split bearings are used (figure 3).

CAP

TOP SHELL

OIL GROOVE

BOTTOM SHELLSUPPORT OR

HOUSING

SHAFT

LUBRICATOR

BEARING

OIL FILM

Figure 3: a split bearing

When the bearing wears, the bearing housing can be separated by removing the two

nuts. The bearing shells can then be removed and replaced. Notice that the inside of

the shells has a groove. This groove is normally fed by a reservoir of oil, which helps

to lubricate the shaft and bearing, thus reducing friction. A car big end is a common

example.

Ball-and-roller bearings Ball and roller bearings change the action of rubbing to that of rolling. Ball and roller

bearings use hardened steel balls or rollers, which rotate inside an inner and outer case.

The outer case or ‘race’ presses into a housing; the inner race is a press fit on the shaft.

These bearings are used in high-speed, high-force applications.


OUTER RACE

INNER RACE

BALL BEARING

CAGE

Ball Thrust ball Roller Needle roller

Figure 4: ball and roller bearings

Clutches

We want to reduce friction in moving parts. To achieve this bearings are used, surface

contact area is minimised and lubricants are used. However, without friction between

the tyres and road, cars would not be able to stay on the roads or even start to move.

Clutches are devices that allow two rotating shafts to be connected and disconnected.

There are two types of clutch, the positive clutch and the friction clutch. A dog clutch

is a positive clutch. This has four interlocking blocks (dogs) on one shaft that can be

interlocked with four dogs on the other shaft.

Figure 1: a dog clutch

When the clutch is engaged, the two dogs are interlocked and the drive shaft rotates

the driven shaft. When the clutch is disengaged, the two shafts are separated. In clutch

systems, the two shafts must be carefully aligned.

Positive-drive clutches require the drive shaft to be stationary when the two clutch

plates are brought together. Friction clutches can be engaged and disengaged while


85

both shafts are still turning. Friction clutches rely on the friction between the plates to

transmit the power from one shaft to another. Figure 2 shows a simple example of a

friction clutch. ENGAGED

DRIVER SHAFTIN MOTION

DRIVEN SHAFTSTATIONARY

Figure 2: a simple friction clutch

Figure 3 shows a multi-plate system used for large transmission forces or limited-

space applications.

Figure 3: a multi-plate clutch


Couplings: task 1

1. Why are couplings used in mechanical systems?

2. What term is used in making sure that a coupling has a positive drive?

3. When alignment is a problem in shafts what mechanical device can be fitted?

4. Why can friction be a problem in mechanical devices?

5. What consequences may occur if friction is not overcome?

6. Thrust bearings are often used in rotating systems. What is the main advantage in

using this type of bearing?

7. If a bearing is required in the middle of a long shaft, how can the problem of

changing it be overcome?

8. How does lubrication work within the bearing housing?

9. Some shafts require to run at high speeds with limited friction. How can this be

done?

10. Explain the term ‘dog clutch’.


87

Couplings: task 2 In the printed version of these materials, issued to Scottish schools in August 2001, this page contained






questions from Principles of Science (Friction) in an interactive way.

1. Friction is a force that tries to resist _ _ _ _ _ _.

2. Ball bearings are used in a dentist’s drill. How does this work?

3. How does friction assist in stopping a bicycle?

4. Friction is a main feature in a car’s breaking system. Explain what part friction

has in assisting both front and rear wheels to stop.

5. Name and explain a situation where friction assists in a car’s road holding

capabilities.

6. When the front brakes of a car are applied, friction occurs. What effects are

produced and how can these be reduced?

7. How does friction affect the clutch of a car?

8. How do bearings assist in the operation of a road repairer’s pneumatic drill?


Documents

Mechanical Systems (part 1)